Question

Find the values of k if the distance between the points (3,k) and (-3,2) is 2√10 units.​

Answer 1

(0 or 4)

Explanation:

The distance formula is

`(x_(2) - x _(1)) {}^(2) + ( {y}^{} _(2) - y _(1)) {}^(2) = {d}^(2)`

Where d is the distance.

So

`( - 3 - 3) {}^(2) + (2 - k) {}^(2) = (2 √(10) ) {}^(2)`

`( - 6) {}^(2) + (2 - k) {}^(2) = 40`

`(2 - k) {}^(2) - 4 = 0`

Let

`u = (2 - k)`

`{u}^(2) - 4 = 0`

We get two solutions,

`(u + 2)(u - 2) = 0`

`(u + 2) = 0`

`u = - 2`

or

`u - 2 = 0`

`u = 2`

SUBSITUTE 2-k back in

`2 - k = - 2`

`k = 4`

or

`2 - k = 2`

`k = 0`